Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $t = \dfrac{r^2 + 6r - 7}{3r^2 + 6r - 9} \times \dfrac{-5r - 15}{r + 9} $
Answer: First factor out any common factors. $t = \dfrac{r^2 + 6r - 7}{3(r^2 + 2r - 3)} \times \dfrac{-5(r + 3)}{r + 9} $ Then factor the quadratic expressions. $t = \dfrac {(r - 1)(r + 7)} {3(r - 1)(r + 3)} \times \dfrac {-5(r + 3)} {r + 9} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac { (r - 1)(r + 7) \times -5(r + 3)} { 3(r - 1)(r + 3) \times (r + 9)} $ $t = \dfrac {-5(r - 1)(r + 7)(r + 3)} {3(r - 1)(r + 3)(r + 9)} $ Notice that $(r - 1)$ and $(r + 3)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {-5\cancel{(r - 1)}(r + 7)(r + 3)} {3\cancel{(r - 1)}(r + 3)(r + 9)} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $t = \dfrac {-5\cancel{(r - 1)}(r + 7)\cancel{(r + 3)}} {3\cancel{(r - 1)}\cancel{(r + 3)}(r + 9)} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $t = \dfrac {-5(r + 7)} {3(r + 9)} $ $ t = \dfrac{-5(r + 7)}{3(r + 9)}; r \neq 1; r \neq -3 $